See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When a ketone is treated with NaNO2/HCl (nitrous acid, HNO2, generated in situ), the reaction is a nitrosation at the alpha-carbon. The initial product is an alpha-nitroso compound, which tautomerizes to the more stable oxime form (=N—OH). Step 1: Identify the starting material. The ketone is CH3CH2—C(=O)—CH3, which is methyl ethyl ketone (butан-2-one). It has two sets of alpha-carbons: the CH3 group (alpha on one side) and the CH2 group (alpha on the other side). Step 2: Determine which alpha position is nitrosated preferentially. In nitrosation with HNO2, the reaction proceeds via the enol tautomer. The more substituted or more accessible enol is attacked. However, the key factor here is that nitrosation preferentially occurs at the less hindered alpha position and yields the more stable product. The methyl group alpha (giving CH3—C(=N—OH)—C(=O)—CH3) involves the terminal methyl on the left side of the carbonyl. Step 3: Mechanism. HNO2 (from NaNO2 + HCl) acts as an electrophile (NO+). It attacks the enol of the ketone at the alpha carbon. The ketone CH3CH2—C(=O)—CH3 can form an enol at the CH3 end: CH3CH2—C(OH)=CH2. Nitrosation here gives CH3CH2—C(=O)—CH2—NO, which tautomerizes to CH3CH2—C(=O)—CH=N—OH (option b). Alternatively, enol formation at the CH2 end: CH3CH=C(OH)—CH3. Nitrosation gives CH3CH(NO)—C(=O)—CH3 (option a), or after tautomerization CH3—C(=N—OH)—C(=O)—CH3 (option c). Step 4: Regioselectivity. Nitrosation preferentially occurs at the more substituted alpha carbon because the more substituted enol is more stable and forms preferentially (Bredt/thermodynamic enol control). The CH2 alpha position (between CH3CH2 and C=O) gives the more substituted enol (trisubstituted double bond) compared to the terminal CH3. The enol CH3CH=C(OH)—CH3 is more substituted (disubstituted alkene) than CH3CH2—C(OH)=CH2 (monosubstituted). Nitrosation of this more stable enol at the alpha CH (the CH3CH= side) gives, after tautomerization, the oxime CH3—C(=N—OH)—C(=O)—CH3, which is option (c). Step 5: The nitroso intermediate CH3—CH(NO)—C(=O)—CH3 (option a) is unstable and tautomerizes to the oxime CH3—C(=N—OH)—C(=O)—CH3 (option c). This oxime is the thermodynamically stable product. Why other options fail: - Option (a): The nitroso tautomer is less stable than the oxime; it is not the major isolated product. - Option (b): This would arise from nitrosation at the less substituted terminal CH3 (CH2 of the ethyl group), which forms the less stable enol; not the major product. - Option (d): This would be the oxime of methyl ethyl ketone at the ketone carbon itself (not an alpha-nitrosation product) — this structure does not arise from this reaction pathway. Therefore, the correct answer is C.